GCE
Statistics (MEI)
Advanced Subsidiary GCE
Unit G241: Statistics 1 (Z1)
Mark Scheme for June 2012
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of the examination. It shows the basis on which marks were awarded by examiners. It does not
indicate the details of the discussions which took place at an examiners’ meeting before marking
commenced.
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candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
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G241
Mark Scheme
Annotations
Annotation in scoris
and 
BOD
FT
ISW
M0, M1
A0, A1
B0, B1
SC
^
MR SC SC this year as MR symbol
missing
Highlighting
Other abbreviations in mark scheme
E1
U1
G1
M1 dep*
cao
oe
rot
soi
www
awrt
Meaning
Benefit of doubt
Follow through
Ignore subsequent working
Method mark awarded 0, 1
Accuracy mark awarded 0, 1
Independent mark awarded 0, 1
Special case
Omission sign
Misread
Meaning
Mark for explaining
Mark for correct units
Mark for a correct feature on a graph
Method mark dependent on a previous mark, indicated by *
Correct answer only
Or equivalent
Rounded or truncated
Seen or implied
Without wrong working
Answer which rounds to
1
June 2012
G241
Mark Scheme
June 2012
Subject-specific Marking Instructions
a.
Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full
marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded.
For subsequent marking you must make it clear how you have arrived at the mark you have awarded.
b.
An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed
to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be
judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working
must always be looked at and anything unfamiliar must be investigated thoroughly.
Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such
work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks
according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved)
you should contact your Team Leader.
c.
The following types of marks are available.
M
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks
are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to
indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in
hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M
mark may be specified.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the
associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded.
B
Mark for a correct result or statement independent of Method marks.
E
A given result is to be established or a result has to be explained. This usually requires more working or explanation than the
establishment of an unknown result.
2
G241
Mark Scheme
June 2012
Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is
ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a
candidate passes through the correct answer as part of a wrong argument.
d.
When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically
says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular
mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has
once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the
other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must
be given.
e.
The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results.
Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are
not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a
solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in
the mark scheme rationale. If this is not the case please consult your Team Leader.
Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow
through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown
within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-byquestion.
f.
Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise.
Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the
norm. Small variations in the degree of accuracy to which an answer is given (eg 2 or 4 significant figures where 3 is expected)
should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a
mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in
the mark scheme rationale. If in doubt, contact your Team Leader.
g.
Rules for replaced work
If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should
do as the candidate requests.
If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last
(complete) attempt and ignore the others.
3
G241
Mark Scheme
June 2012
NB Follow these maths-specific instructions rather than those in the assessor handbook.
h.
For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered,
mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally
appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question.
Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
4
G241
Question
1 (i)
Mark Scheme
Answer
P(All blue) =
June 2012
Marks
30 29 28
= 0.2071
 
50 49 48
M1
For
30
 (as part of a
50
triple product)
Guidance
(30/50)3 = 0.216 scores
M1M0A0
k
( k 1) ( k  2)


for values of k
50
49
48
other than 30 scores M1M0A0
Zero for binomial unless
simplifies to (3/5)3
OR
 30   50 
 / 
3 3
M1
For product of other two
fractions
A1
CAO
= 4060/19600 = 29/140 = 0.2071
SC2 for P(All red) =
0.0582
M2 for the complete method
1
(ii)
20 19 18
P(All red) =
= 0.0582 or
 
50 49 48
 20   50 
 / =
3 3
0.0582
P(At least one of each colour)
= 1 – (0.2071 + 0.0582) = 0.7347
[3]
M1
For P(All red)
M1
For 1 – (0.2071 + 0.0582)
A1
[3]
CAO
Correct working but then
multiplied or divided by some
factor scores M1M0A0
Accept 0.21 with working and
0.207 without working
Allow unsimplified fraction as
final answer 24360/117600 oe
SC2 for 1 – (30/50)3 – (20/50)3
= 1 – 0.216 – 0.064 = 0.72,
providing consistent with (i) . If
not consistent with (i) M0M0A0
260
13 36
 29 57 
or 1  

 1

  1
980
49 49
 140 980 
OR
P(2b,1r)+P(1b,2r)
(M1)
5
30 29 20
 
50 49 48
20 19 30
or
 
50 49 48
For either
Allow 0.73 with working
Allow unsimplified fraction as
final answer 86400/117600 oe
Allow M1 for
3×(30/50)2×(20/50) or
3×(30/50) ×(20/50) 2 and second
M1 for sum of both if = 0.72
If not consistent with (i)
M0M0A0
G241
Mark Scheme
Question
2
2
(i)
(ii)
Marks
(M1)
Answer
= 3
June 2012
30 29 20
20 19 30
  + 3  
50 49 48
50 49 48
  
50 49  48 
= 3 × 0.1480 + 3 × 0.0969 = 0.7347
(A1)
OR
Either
(M1)
9
 30   20   50 
   /  
2 1 3
 30   20   50 
  /  
1 2 3
or 
(M1)
(A1)
M1
M1
C3 × 5C3 = 84 × 10 = 840
A1
[3]
M1
M1
Total number of ways of answering 6 from 14 is 14C6 = 3003
Probability =
Guidance
For sum of both or for 3× NB M2 also for
either
30 20  48 
840
40

= 0.27972 = 0.280
3003 143
A1
CAO
For sum of both
CAO
For either 9C3 or 5C3
For product of both
correct combinations
CAO
even if not multiplied by 3
Allow 0.73 or better with
working
Zero for permutations
For 14C6 seen in part (ii)
For their 840/ 3003 or
their 840/14C6
FT their 840
Allow full marks for
unsimplified fractional answers
(M1)
For product of fractions
(M1)
(A1)
For 6C3× correct product
SC1 for 6C3 × (5/14)3 × (9/14)3 =
0.2420
[3]
OR
6
C3 × 5/14 × 4/13 × 3/12 × 9/11 × 8/10 × 7/9 = 0.280
6
G241
Question
3 (i)
Mark Scheme
Marks
M1
Answer
X ~ B(30, 0.85)
M1
 30 
 × 0.8529 × 0.151 = 30 × 0.0013466 = 0.0404
29
 
P(X = 29) = 
3
3
(ii)
(iii)
June 2012
P(X = 30) = 0.8530 = 0.0076
P(X ≥ 29) = 0.0404 + 0.0076 = 0.0480
For 0.85 × 0.15 =
0.0013466
 30 
 × p29 × q1
29
 
For 
CAO
M1
M1
For 0.8530
For P(X = 29) + P(X =
30) (not necessarily
correct, but both attempts
at binomial, including
coefficient in (i))
CAO
[2]
7
1
A1
[3]
A1
[3]
M1
A1
Expected number = 10 × 0.0480 = 0.480
Guidance
29
For 10 × their (ii)
FT their (ii) but if answer
to (ii) leads to a whole
number for (iii) give
M1A0
With p + q = 1
Allow 0.04 www
If further working (EG P(X=29)
–P(X=28)) give M2A0
Allow eg 0.04+0.0076=0.0476
Allow 0.05 with working
provided (ii) between 0 and 1
Do not allow answer rounded to
0 or 1.
G241
Mark Scheme
June 2012
Marks
M1
M1
Question
Answer
2
(A)
P(third selected) = 0.92 × 0.08 = 0.0677
4 (i)
Or = 1058/15625
A1
[3]
4
(i)
4
(ii)
(B)
P (second) + P(third)
= (0.92 × 0.08) + (0.922 × 0.08)
= 0.0736 + 0.0677 = 0.1413
= 2208/15625
P(At least one of first 20) = 1 - P(None of first 20)
M1
A1
[2]
M1
= 1 – 0.9220 = 1 – 0.1887 = 0.8113
M1
A1
[3]
8
Guidance
2
For 0.92
For p2 × q
With p + q = 1
With no extra terms
Allow 0.068 but not 0.067 nor
0.07
SC1 for ‘without replacement’ method 92/100×91/99×8/98
=0.0690
For 0.92 × 0.08
With no extra terms
FT their 0.0677
Allow 0.141 to 0.142 and
allow 0.14 with working
SC1 for answer of 0.143 from ‘without replacement’ method
0.9220
Accept answer of 0.81 or better
from P(1) + P(2) + ... , or SC2 if
all correct working shown but
wrong answer
No marks for ‘without
replacement’ method’
1 – 0.9220
CAO
Allow 0.81 with working but
not 0.812
CAO
G241
Question
5
Mark Scheme
June 2012
Marks
B1
Answer
Let p = probability that a randomly selected frame is faulty
H0: p = 0.05
B1
H1: p > 0.05
P(X ≥ 4)
B1
B1
= 1- P(X ≤ 3) = 1 – 0.9891 = 0.0109
B1*
9
Guidance
For definition of p in context
Minimum needed for B1 is p = probability that frame/bike is
faulty. Do not allow is p = probability that it is faulty
Allow p = P(frame faulty)
Definition of p must include word probability (or chance or
proportion or percentage or likelihood but NOT possibility).
Preferably as a separate comment. However can be at end of
H0 as long as it is a clear definition ‘ p = the probability that
frame is faulty, NOT just a sentence ‘probability is 0.05’
Do NOT allow ‘p = the probability that faulty frames have
increased’
H0 : p(frame faulty) = 0.05, H1 : p(frame faulty) > 0.05 gets
B0B1B1
Allow p=5%, allow θ or π and ρ but not x. However allow
any single symbol if defined
Allow H0 = p=0.05, Allow H0 : p=1/20
Do not allow H0 : P(X=x) = 0.05, H1 : P(X=x) > 0.05
Do not allow H0: =0.05, =5%, P(0.05), p(0052), p(x)=0.05,
x=0.05 (unless x correctly defined as a probability)
Do not allow H1:p≥0.05,
Do not allow H0 and H1 reversed
Allow NH and AH in place of H0 and H1
For hypotheses given in words allow Maximum B0B1B1
Hypotheses in words must include probability (or chance or
proportion or percentage) and the figure 0.05 oe.
For notation P(X ≥ 4) or
1- P(X ≤ 3)
This mark may be
implied by 0.0109 as long
as no incorrect notation.
For 0.0109, indep of
previous mark
No further marks if point probs
used - P(X = 4) = 0.0094
DO NOT FT wrong H1
But if H1 is p ≥ 0.05 allow the
rest of the marks if earned so
max 7/8
Or for 1 – 0.9891
G241
Mark Scheme
Question
June 2012
Marks
M1*
dep
A1*
Answer
0.0109 < 0.05
So reject H0
There is evidence to suggest that the proportion of faulty frames has
increased.
E1*
Dep on
A1
Guidance
For comparison with 5%
or significant or ‘accept
H1’
Must include ‘sufficient evidence’ or something similar such
as ‘to suggest that’ ie an element of doubt for E1. ‘Sufficient
evidence’ or similar can be seen in the either the A mark or
the E mark.
[8]
OR Critical region method:
Let X ~ B(18, 0.05)
P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.9419 = 0.0581 > 5%
(B1)
For 0.0581
P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – 0.9891 = 0.0109 < 5%
(B1)
For 0.0109
(M1)
For at least one correct
comparison with 5%
CAO for critical region
and significant oe
(A1)
So critical region is {4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}
4 lies in the critical region, so significant,
There is evidence to suggest that the proportion of faulty frames has
increased.
10
(E1)
No marks if CR not justified
Do not insist on correct notation
as candidates have to work out
two probabilities for full marks
Condone {4,5 ..}, X ≥ 4, oe but
not P(X ≥ 4)
G241
Question
6 (i)
Mark Scheme
Engine size
500 ≤ x ≤ 1000
1000 < x ≤ 1500
1500 < x ≤ 2000
2000 < x ≤ 3000
3000 < x ≤ 5000
Answer
Frequency Group width
7
500
22
500
26
500
18
1000
7
2000
Frequency density
0.014
0.044
0.052
0.018
0.0035
June 2012
Marks
M1
Guidance
At least 4 fds correct for M1
M1 can be also be gained from freq per 1000 – 14, 44, 52,
18, 3.5 (at least 4 correct) and A1 for all correct
or freq per 500 - 7, 22, 26, 9, 1.75
Accept any suitable unit for fd, eg freq per 1000, BUT NOT
FD per 1000
Allow fds correct to at least three dp
If fd not explicitly given, M1 A1 can be gained from all
heights correct (within one square) on histogram (and M1A0
if at least 4 correct)
Allow restart with correct heights if given fd wrong
A1
G1(L1)
For fd’s all correct
linear scales on both axes and label on vertical axis
Label required on vert axis IN RELATION to first M1 mark
ie fd or frequency density or if relevant freq/1000, etc (NOT
fd/1000, but allow fd×1000, etc)
Accept f/w or f/cw (freq/width or freq/class width)
Ignore horizontal label and allow horizontal scale to start at
500
Can also be gained from an accurate key
INCORRECT DIAGRAMS:
Frequency diagrams can get M0, A0, G0, G1, G0 MAXIMUM
Thus frequency density = frequency × width, frequency/midpoint etc
gets MAX M0A0G0G1G0
Frequency polygons MAX M1A1G0G0G0
11
G1(W1)
Width of bars
Must be drawn at 500, 1000etc NOT 499.5 or 500.5 etc NO
GAPS ALLOWED
Must have linear scale.
No inequality labels on their own such as 500≤S<1000, etc
but allow if a clear horizontal linear scale is also given.
G241
Mark Scheme
Question
6
6
(ii)
(iii)
Marks
G1(H1)
Answer
Do not know exact highest and lowest values so cannot tell what the
midrange is.
OR
No and a counterexample to show it may not be 2750
OR
(500 + 5000) / 2 = 2750. But very unlikely to be absolutely correct but
probably close to the true value. Some element of doubt needed. Allow
‘Likely to be correct’
Mean =
(750  7)  (1250  22)  (1750  26)  (2500  18)  (4000  7)
80

June 2012
[5]
E1
x2 f  (7502  7)  (12502  22)  (17502  26)  (25002 18)  (40002  7)
s
Allow comment such as 'Highest value could be 5000 and
lowest could be 500 therefore midrange could be 2750'
NO mark if incorrect calculation
[1]
M1
M1
= 3937500 + 34375000 + 79625000 + 112500000 + 112000000
=342437500
Sxx  342437500 -
Height of bars
FT of heights dep on at least 3 heights correct and all must
agree with their fds
If fds not given and one height is wrong then max
M1A0G1G1G0
– visual check only (within one square) –no need to measure
precisely
Sight of 1750 AND 3000 (min and max of midrange) scores
E1
A1
151250
 1891
80
Guidance
For midpoints (at least 3 correct)
No marks for mean or sd unless using midpoints
Answer must NOT be left as improper fraction
CAO
Accept correct answers for mean (1890 or 1891) and sd (850
or 846 or 845.5) from calculator even if eg wrong Sxx given
For sum of at least 3 correct multiples fx2
Allow M1 for anything which rounds to 342400000
151250 2
 56480469
80
56480469
 714943  846
79
A1
Only an estimate since the data are grouped.
E1
indep
[5]
12
Only penalise once in part (iii) for over specification, even if
mean and standard deviation both over specified.
Allow SC1 for RMSD 840.2 or 840 from calculator
Or for any mention of midpoints or ‘don’t have actual data’
or ‘data are not exact’ oe
G241
Question
6 (iv)
6
6
(v)
(vi)
Mark Scheme
June 2012
Marks
M1
Answer
x – 2s = 1891 – (2  846) = 199
Allow 200
x + 2s = 1891 + (2  846) = 3583
Allow 3580 or 3600
A1
So there are probably some outliers
E1
[3]
M1
M1
indep
A1
3
Number of cars over 2000 cm = 25/80 × 2.5 million = 781250
So duty raised = 781250 × £1000 = £781 million
Because the numbers of cars sold with engine size greater than 2000
cm3 might be reduced due to the additional duty.
[3]
E1
[1]
13
Guidance
For either.
FT any positive mean and their positive sd/rmsd for M1
Only follow through numerical values, not variables such as
s, so if a candidate does not find s but then writes here ‘limit
is 40.76+ 2  standard deviation’, do NOT award M1
No marks in (iv) unless using x + 2s or x – 2s
For both (FT)
Do NOT penalise over specification here as it is not the final
answer
Must include an element of doubt
Dep on upper limit in range 3000 – 5000
Allow comments such as ‘any value over 3583 is an outlier’
Ignore comments about possible outliers at lower end.
For 25/80× 2.5 million or (18+7) /80× 2.5 million
For something × £1000 even if this is the first step
CAO
NB £781250000 is over specified so only 2/3
Allow any other reasonable suggestion
Condone ‘sample may not be representative’
Allow ‘sample is not of NEW cars’
G241
Mark Scheme
Question
7 (i)
P(X = 0) = 0.4 × 0.5 = 0.025
7
P(X = 1) = (0.6 × 0.54) + (4 × 0.4 × 0.5× 0.53)
(ii)
4
= 0.0375 + 0.1 = 0.1375
June 2012
Marks
M1
A1
[2]
M1*
Answer
NB ANSWER GIVEN
M1*
NB ANSWER GIVEN
M1*
dep
A1
[4]
7
Guidance
4
For 0.5
For 0.6 × 0.54 seen as a single term (not multiplied or divided
by anything)
For 4 × 0.4 × 0.54 Allow 4 × 0.025
Watch out for incorrect methods such as (0.4/4)
0.1 MUST be justified
For sum of both , dep on both M1’s
(iii)
G1
For labelled linear scales on both axes
Dep on attempt at vertical line chart. Accept P on vertical
axis
G1
For heights – visual check only but last bar taller than first
and fifth taller than second and fourth taller than third.
Lines must be thin (gap width > line width). All correct.
Zero if vertical scale not linear
Everything correct but joined up tops G0G1 MAX
Everything correct but f poly G0G1 MAX
Everything correct but bar chart G0G1 MAX
Curve only (no vertical lines) gets G0G0
Best fit line G0G0
Allow transposed diagram
[2]
14
G241
Mark Scheme
Marks
E1
[1]
Question
7 (iv)
Answer
‘Negative’ or ‘very slight negative’
7
E(X) = (0×0.025) + (1×0.1375) + (2×0.3) + (3×0.325) + (4×0.175)
+ (5×0.0375)
= 2.6
E(X2) = (0×0.025) + (1×0.1375) + (4×0.3) + (9×0.325) + 16×0.175)
+ (25×0.0375) = 0 + 0.1375 + 1.2 + 2.925 + 2.8 + 0.9375 = 8
Var (X) = 8 – 2.62
(v)
= 1.24
7
(vi)
June 2012
P(Total of 3) = (3×0.325×0.0252) + (6×0.3×0.1375×0.025) +
0.13753 = 3 × 0.000203 + 6 × 0.001031 + 0.002600=
0.000609 + 0.006188 + 0.002600 = 0.00940
(= 3×13/64000 + 6×33/32000 + 1331/512000)
Guidance
M1
A1
E0 for symmetrical
but E1 for (very slight) negative skewness even if also
mention symmetrical
Ignore any reference to unimodal
For Σrp (at least 3 terms correct)
CAO
M1*
For Σr2p (at least 3 terms correct)
M1*
dep
A1
[5]
for – their E( X )²
M1
M1
M1
A1
[4]
FT their E(X) provided Var( X ) > 0
USE of E(X-µ)2 gets M1 for attempt at (x-µ)2 should see (2.6)2, (-1.6)2, (-0.6)2, 0.42, 1.42, 2.42 (if E(X) correct but FT
their E(X)) (all 5 correct for M1), then M1 for Σp(x-µ)2 (at
least 3 terms correct)
Division by 5 or other spurious value at end gives max
M1A1M1M1A0, or M1A0M1M1A0 if E(X) also divided by
5.
Unsupported correct answers get 5 marks.
For decimal part of first term 0.325×0.0252
For decimal part of second term 0.3×0.1375×0.025
For third term – ignore extra coefficient
All M marks above depend on triple probability products
CAO: AWRT 0.0094. Allow 0.009 with working.
NOTE RE OVER-SPECIFICATION OF ANSWERS
If answers are grossly over-specified, deduct the final answer mark in every case. Probabilities should also be rounded to a sensible degree of
accuracy. In general final non probability answers should not be given to more than 4 significant figures. Allow probabilities given to 5 sig fig. In
general accept answers which are correct to 3 significant figures when given to 4 or 5 significant figures.
If answer given as a fraction and as an over-specified decimal – ignore decimal and mark fraction.
15
G241
Mark Scheme
June 2012
ADDITIONAL NOTES RE Q5
Comparison with 95% method
If 95% seen anywhere then
B1 for P(X ≤ 3)
B1 for 0.9891
M1* for comparison with 95% dep on B1
A1* for significant oe
E1*
Smallest critical region method:
Either:
Smallest critical region that 4 could fall into is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} gets B1 and has size 0.0109 gets B1, This is < 5% gets
M1*, A1*, E1* as per scheme
NB These marks only awarded if 4 used, not other values.
Use of k method with no probabilities quoted:
P(X ≥ 3) = 1 – P(X ≤ 2) > 5%
P(X ≥ 4) = 1 – P(X ≤ 3) < 5%
These may be seen in terms of k or n.
Either k = 4 or k – 1 = 3 so k = 4 gets SC1
so CR is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} gets another SC1and conclusion gets another SC1
Use of k method with one probability quoted:
1 - 0.9891 < 5% or 0.0109 < 5% gets B0B1M1
P(X ≤ k - 1) = P(X ≤ 3)
so k – 1 = 3 so k = 4 (or just k = 8 )
so CR is {4,5,6,7,8,9,10,11,12,13,14,15, 16, 17, 18} and conclusion gets A1E1
16
G241
Mark Scheme
June 2012
Two tailed test done but with correct H1: p>0.05
Hyp gets max B1B1B1
if compare with 5% ignore work on lower tail and mark upper tail as per scheme but withhold A1E1
if compare with 2.5% no marks B0B0M0A0E0
Line diagram method
B1 for squiggly line between 3 and 4 or on 4 exclusively (ie just one line), B1dep for arrow pointing to right, M1 0.0109 seen on diagram from
squiggly line or from 4, A1E1 for correct conclusion
Bar chart method
B1 for line clearly on boundary between 3 and 4 or within 4 block exclusively (ie just one line), B1dep for arrow pointing to right, M1 0.0109 seen
on diagram from boundary line or from 8, A1E1 for correct conclusion.
Using P(Not faulty) method
H0: p=0.95, H1: p<0.95 where p represents the prob that a frame is faulty gets B1B1B1.
P(X≤14)=0.0109 < 5% So significant, etc gets B1B1M1A1E1
NB
If H0: p=0.5, H1: p>0.5, etc seen, but then revert to 0.05 in working allow marks for correct subsequent working. However if 0.5 used consistently
throughout, then max B1 for definition of p and possibly B1 for notation P(X ≥ 4).
17
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GCE
Statistics (MEI)
Advanced Subsidiary GCE AS H132
OCR Report to Centres
June 2012
Oxford Cambridge and RSA Examinations
OCR Report to Centres – June 2012
G241 Statistics 1
General Comments
The level of difficulty of the paper appeared to be appropriate and there was no evidence of
candidates being unable to complete the paper in the allocated time. Many candidates were well
prepared but a number struggled to handle even the more standard parts of questions. Most
candidates supported their numerical answers with appropriate explanations and working.
It is pleasing to report that the hypothesis test question was generally answered very well, with
most candidates not only giving their hypotheses in terms of p but also defining p as the
probability of a bike frame being faulty. Most candidates also included an element of doubt in
their conclusion saying eg. ‘There is sufficient evidence to suggest that the proportion of faulty
frames has increased’. Unfortunately most candidates lost marks due to over specification of
some of their answers, despite recent examiners’ reports warning against this. Particular
examples are given in the comments on 6(iii) and 6(v) below.
The comments below on individual questions are based on candidate responses to both 4766
and G241.
Comments on Individual Questions
1) (i)
The majority of candidates gained full marks in this part. However a significant number
completed this question as if the bulbs were replaced before each choice. This was
given 1 mark out of the 3 available.
1) (ii)
Once again the majority of candidates gained at least 2 marks out of 3. Those who
answered by finding the P(2 blue and one red) and adding it to P(2 red and one blue)
were in the majority, but were less successful than those who found
1 – (P(3 red) + P(3 blue)). This was due to the omission of the 3 possible
arrangements of each probability.
2) (i)
Again the majority of candidates gained full marks. A fairly common error was to add
rather than multiply 9C3 and 5C3. A small number of candidates tried to use
permutations rather than combinations.
2) (ii)
Many candidates gained full credit for dividing their answer to part (i), correct or not, by
3003. Those who did not see the connection with part (i) did not fare so well, and even
if they found the correct product of fractions they rarely multiplied this by 6C3.
3) (i)
This question was very well answered, with most candidates scoring all 3 marks.
3) (ii)
Many fully correct responses were seen, although a number of candidates calculated
P(X<29) or P(X≤29) rather than P(X≥29).
3) (iii)
Many candidates gained full credit here, even if as a follow through from their answer
to part (ii). A common error was to multiply their answer to part (ii) by 30 or 300 instead
of by 10. A number of candidates also rounded their answer to a whole number,
thereby losing the second mark.
4) (i)(A) Most candidates scored full marks, but a significant number scored zero. Candidates
needed to multiply 0.922 by 0.08, but a significant number simply worked out 0.083,
which gained no credit.
1
OCR Report to Centres – June 2012
4) (i)(B) In this part candidates needed to multiply 0.92 by 0.08 then add this product to their
answer to part (i). This was often achieved successfully, but a number of candidates
gave their answer to 6 significant figures, thus losing the second mark.
4) (ii)
Many fully correct responses were seen. Most other candidates gained no credit as
they had no real idea of what was required.
5)
Many candidates were awarded at least 7 out of the 8 marks available. The
hypotheses were generally correct and well defined but a minority of candidates still
omitted a definition of p. Only a small number of candidates incorrectly used point
probabilities. Many candidates used the first method in the scheme, usually
successfully. A smaller number used the critical region method, again fairly
successfully, but a number thought that the critical region started at 3. Some
candidates who used the critical region method, failed to justify their critical region. In
this case they were only eligible for the first 3 marks for the hypotheses.
6) (i)
On the whole, this question was answered well, with a very high proportion of
candidates calculating the frequency densities correctly. Of those candidates who did
not calculate the FD correctly, most achieved 1 mark for the correct widths. There
were very few inequality labels on the x axis. However, candidates should be reminded
that they need to label the vertical axis. Drawing of the bars was done well although a
few candidates struggled to draw a bar of height 0.0035.
6) (ii)
Many candidates thought that the calculation involved subtraction rather than addition
and even when the calculation was correct, there was often no element of doubt to
their conclusion.
6) (iii)
On the whole this question was very well answered. It was extremely common to
award 4 marks in total, due to the over specification of answers. Many candidates
gave the exact answer 1890.625, but an element of sensible rounding, to say 1891 or
even 1890, was looked for. A significant number did not find the standard deviation
correctly, sometimes giving the root mean square deviation or calculating (fx)2 rather
than fx2. The explanation mark was very well answered.
6) (iv)
Again this was also very well answered. Even candidates who had made errors in the
previous part usually gained follow through marks. Most candidates knew that the
limits for outliers were mean ± 2 standard deviations. A number of candidates did not
include an element of doubt in their conclusion about the number of outliers and thus
were not awarded the final mark.
6) (v)
The modal mark in this question was 2 out of 3, scored by approximately 2/3 of the
candidature. This was due to candidates over specifying their answer, giving it as
781250000 rather than for example 780000000. Candidates who were unsure how to
do this part nevertheless usually gained a method mark for multiplying by 1000.
6) (vi)
Where candidates achieved this mark, they often realised that the duty would reduce
the sales of larger cars. They also achieved this mark where they stated that the
sample may not be representative, although this needed to be very clearly stated for
the mark to be awarded. A number of candidates erroneously stated that people would
refuse to pay the duty.
7) (i)
This part was very well answered.
2
OCR Report to Centres – June 2012
7) (ii)
Again many fully correct responses were seen. Many other candidates scored 1 mark
out of 4 for finding 0.6 x 0.54 = 0.0375. Some candidates multiplied 0.6 x 0.54 or
0.4 x 0.54 or indeed both by 5C1 rather than by 4C1.
7) (iii)
Approximately half of candidates scored full marks in this part. However many lost one
or both marks for a number of errors – a non linear vertical scale, one or both labels
missing, heights incorrect (particularly the final height), or less often a frequency
polygon or a point plot. Candidates should be advised to use a ruler in questions such
as this.
7) (iv)
Again approximately half of candidates scored the mark here. The most popular
answer was ‘slightly negative’, but some said positive skew or symmetrical and/or
unimodal.
7) (v)
There were very many correct answers, with even the weakest candidates often
scoring full marks. Arithmetic errors were common often because of writing the
probabilities incorrectly as eg 0.375 rather than 0.0375. Only a few candidates left the
variance as 8 or did not square E(X). Very few incorrectly divided by 5, unlike in
previous sessions.
7) (vi)
Candidates needed to have a very good understanding of probability to gain marks in
this part. However, some got 1, 2 or 3 products of probabilities correct but very few
had the coefficients correct.
3